What is the purpose of taking coefficients as 1 in numerical solutions? How can we recover the real solution after getting a solution by solving with parameters set to 1?
For example, on my case to solve the Shrödinger equation via finite difference method, the author took the coefficients like $h$ and $2m$ as 1, and gets result with this form.

How do the result of this new equation relate to the solution for the original one?
 A: Another way to think of it is that you're dividing the equation by $\frac{\hbar}{2m}$. Any solution to 
$$ \hat{H} \psi = 0 $$
is a solution to 
$$ \frac{2 m \hat{H}}{\hbar} \psi = 0 .$$
Since the second equation is just the first multiplied by something on both sides. You are interested in finding the space of solutions of the differential equation, so either one is okay. In a general differential equation, you are right, you can't necessarily just set things to one and get the same set of solutions. 
A: The system is called atomic units. It is to simplify the equations, both notationally and computationally. However, some care is required in derivation, because it is no longer possible to perform sanity checks by dimensional analysis.
Now usually, one assumes m=1, instead of 2m=1. With m=1, the energy unit is called Hartrees and it is about 27 eV. When 2m=1 as in your case, the energy unit is called Rydberg and it is twice the amount of Hartrees, about 13eV. The total energy of hydrogen atom is -0.5 Hartrees or -1 Rydbergs. More info can be found in wikipedia.
If you want comversion factors, just find any combination of hbar, electron mass etc In SI units, which produce the desired derived unit. It might not be trivial since the basic units have dimensions like Js. See derived atomic units of https://en.m.wikipedia.org/wiki/Atomic_units for more details.
A: Consider the time independent Schrodinger equation (in the position basis)
$$\left[ \left( \frac{-\hbar^2}{2m} \right) \left( \frac{d}{dx} \right)^2 + V(x) \right] \Psi(x) = E \Psi(x) \, .$$
In order to solve this numerically, we have to turn it into an equation with just numbers and no physical dimensions like length and mass.
There are essentially two ways to do this.
Implied dimensions
Choose a system of units and express everything in that system.
In that case, we could choose SI units and pick the values for $m$, $\hbar$, etc. for that unit system.
For example, the value of $\hbar$ would be $1.05 \times 10^{-34}$.
If we do this, then each term in the equation would implicitly have dimensions of energy, and in fact would have units of Joules since we're using the SI system.
This is not a great strategy because you're working with really large and really small numbers.
In particular, this makes it difficult to know things like how finely you need to divide the position axis to expect to get meaningful results.
Dimensionless
The better solution is to remove all dimensions from the problem.
First, note that we can discretize the second derivative and turn it into a matrix operator
$$
\frac{1}{a^2} \left[ \begin{array}{ccccc}
\ddots &&  &&  &&  && \\
&& -2 && 1 && 0 && \\
&& 1 && -2 && 1 && \\
&& 0 && 1 && -2 && \\
&&  &&  &&  && \ddots \end{array}
\right]
\equiv \frac{1}{a^2} T
$$
where $a$ is the distance between each discrete point and $T$ is the dimensionless matrix.
With this construction, the Schrodinger equation becomes
$$\left[ \left( \frac{-\hbar^2}{2 m a^2} \right) T + V(x) \right] \Psi(x) = E \Psi(x) \, .$$
Denoting $E_0 \equiv \hbar^2 / (m a^2)$, we divide through by $E_0$ to get
$$\left[ -\frac{1}{2} T + \frac{V(x)}{E_0} \right] \Psi(x) = \frac{E}{E_0} \Psi(x) \, .$$
Perhaps this is even more clear if we just define some new dimensionless parameters
$$\epsilon \equiv E/E_0 \qquad v(x) \equiv V(x)/E_0$$
such that the Schrodinger equation is dimensionless:
$$\left[ -\frac{1}{2}T + v(x) \right] \Psi(x) = \epsilon \Psi(x) \, . \tag{1}$$
I think this is a much better way to explain what's going on rather than to say we've "set $\hbar$ and $m$ to 1".
Yet another way to say all this is that we're working with $(H/E_0)\Psi = (E/E_0)\Psi$ instead of the original $H \Psi = E \Psi$.
Let's now go back to the original question:

How can we recover the real solution after getting a solution by solving with parameters set to 1?

If we numerically find the eigenvalues of the dimensionless equation $(1)$, we've found values of $\epsilon$.
In the original problem we want to know the eigenvalues $E$, so just multiply the values of $\epsilon$ by $E_0$.
